Andrew Conti agconti

Get a PGP key from Keybase.

A walkthrough on how to get a pgp key from keybase.

Go to keybase and create an account.

I hereby claim:

I am agconti on github.
I am agconti (https://keybase.io/agconti) on keybase.
I have a public key whose fingerprint is F98C 1E59 630E F558 ED5B D1D2 62FD E34C 2634 4083

To claim this, I am signing this object:

I hereby claim:

To claim this, I am signing this object:

	layout python3

	# Automatically install dependencies when requirements.txt changes
	LAST_INSTALLED_DEPENDENCIES_FLAG="$VIRTUAL_ENV/.dependencies_installed"
	echo $LAST_INSTALLED_DEPENDENCIES_FLAG

	if [ ! -f $LAST_INSTALLED_DEPENDENCIES_FLAG ] \|\| [ requirements.txt -nt $LAST_INSTALLED_DEPENDENCIES_FLAG ]; then
	echo "Installing dependencies..."
	pip install -r requirements.txt
	touch $LAST_INSTALLED_DEPENDENCIES_FLAG

	<html>
	<body>
	<div data-ad="true" class="reactive-embed" data-replay-id="ef3a1887-e996-4e85-8ab0-368fdab0327d"></div>
	<script src="https://consumer-app-staging.reactive.live/bundle.js"></script>
	</body>
	</html>

	const { promisify } = require('util')
	const { promises: fs } = require('fs')
	const hostile = require('hostile')
	const setHost = promisify(hostile.set)
	const removeHost = promisify(hostile.remove)
	const LOCALHOST = '127.0.0.1'
	const hosts = [
	[LOCALHOST, 'youralias'],
	]

	class Solution_recursive:
	"""
	Runtime: O(2^n)
	Spacetime: O(1)

	The classic approach to solving the Fibonacci sequence. The code is elegant but its runtime isn't
	very pretty. Its recursive nature necessitates that we calculate some values of f(n) multiple times.
	Consider this call graph where duplicated calls are denoted with `~`:

	fib(7)

	const http = require("http")

	const getWeather = () => new Promise((resolve, reject) => {
	const options = {
	hostname: 'nodejs.org',
	path: '/dist/index.json',
	method: 'GET',
	};

	http.get(options, (res) => {